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Consider the PDE (letting $u = u(t,x)$)

$$ u_t + (ab - c^2 - ax)u_x - \frac{c^2}{2}u_{xx} - au = 0, $$ where $a,b,c > 0$. In a paper I'm reading, they note this PDE has been shown to be well-defined for $u \in [0, \infty)$ and give the "stationary" solution $u_*$ ( I believe this means the solution for $t \to \infty$) as

$$ u_*(x) = \frac{\alpha^\alpha}{\Gamma(\alpha)} \frac{x^{\alpha-1}}{b^\alpha}e^{-\alpha x/b}, $$ where $\alpha = \frac{2ab}{c^2}$. This is just the $gamma(\alpha, \frac{b}{\alpha})$ PDF.

They don't give the solution, nor do they derive the stationary solution. I've tried to use integrating factors and changes of variable for the solution, but PDEs are not my strong point. A push in the right direction would be greatly appreciated!

Link to paper: paper. I've rewritten the PDE from the paper (eqn (8) in paper) in more familiar notation, and their stationary solution is just under it (eqn (9)).

bcf
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2 Answers2

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"Stationary solution" means a solution that doesn't depend on $t$. If $u(t,x) = U(x)$ you get the ODE $$ -(ab - c^2 - ax) U'(x) - \dfrac{c^2}{2} U''(x) - a U = 0$$ Your alleged stationary solution $u^*$ does not satisfy this, so there appears to be a mistake somewhere.

In fact, a second-order linear differential equation satisfied by $u_*(x) = x^{\beta} e^{-\gamma x}$ would be $$c x^2 u'' + (b x + 2 c \gamma x^2) u' + ( b (\gamma x - \beta) + c (\gamma x^2 - \beta(\beta-1)) u = 0$$ but that doesn't look much like your equation.

Robert Israel
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  • Okay, there was a slight mistake. The $u_x$ term should have had a $+$ sign, not a $-$ sign. I edited the equation. – bcf Feb 26 '15 at 01:18
  • That still doesn't work. You can see this because if $u_(x) = x^\beta e^{-\gamma x}$, $u_''$ will have a term in $x^{\beta-2} e^{-\gamma x}$ which does not match any term in $u_'$ or $x u_'$ or $u_*$. – Robert Israel Feb 26 '15 at 01:20
  • Hmm. If you're up for it, I provided a link to the paper above. Perhaps I'm missing something. – bcf Feb 26 '15 at 01:24
  • Yes, you are: the factor $v$ in the term $\dfrac{\partial^2}{\partial^2 v} \left[ v \Pi_t(v)\right]$, I think. – Robert Israel Feb 26 '15 at 01:31
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Hint:

Let $\begin{cases}x_1=ab-c^2-ax\\t_1=t\end{cases}$ ,

Then $u_x=u_{x_1}(x_1)_x+u_{t_1}(t_1)_x=-au_{x_1}$

$u_{xx}=(-au_{x_1})_x=(-au_{x_1})_{x_1}(x_1)_x+(-au_{x_1})_{t_1}(t_1)_x=a^2u_{x_1x_1}$

$u_t=u_{x_1}(x_1)_t+u_{t_1}(t_1)_t=u_{t_1}$

$\therefore u_{t_1}-ax_1u_{x_1}-\dfrac{a^2c^2}{2}u_{x_1x_1}-au=0$

$u_{t_1}=\dfrac{a^2c^2}{2}u_{x_1x_1}+ax_1u_{x_1}+au$

With reference to Change variables into Fokker-Planck PDE,

Let $\begin{cases}x_2=x_1e^{at_1}\\t_2=t_1\end{cases}$ ,

Then $u_{x_1}=u_{x_2}(x_2)_{x_1}+u_{t_2}(t_2)_{x_1}=e^{at_1}u_{x_2}=e^{at_2}u_{x_2}$

$u_{x_1x_1}=(e^{at_2}u_{x_2})_{x_1}=(e^{at_2}u_{x_2})_{x_2}(x_2)_{x_1}+(e^{at_2}u_{x_2})_{t_2}(t_2)_{x_1}=e^{2at_2}u_{x_2x_2}$

$u_{t_1}=u_{x_2}(x_2)_{t_1}+u_{t_2}(t_2)_{t_1}=ax_1e^{at_1}u_{x_2}+u_{t_2}$

$\therefore ax_1e^{at_1}u_{x_2}+u_{t_2}=\dfrac{a^2c^2}{2}e^{2at_2}u_{x_2x_2}+ax_1e^{at_2}u_{x_2}+au$

$u_{t_2}-au=\dfrac{a^2c^2}{2}e^{2at_2}u_{x_2x_2}$

Which is separable.

Can you take it from here?

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doraemonpaul
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